University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 24 - Capacitance and Dielectrics - Problems - Exercises - Page 809: 24.2

Answer

a) The capacitance of the capacitor is $=2.65\times10^{-12}F$ b) The voltage or potential difference is $16,400 V$ or $16.4 kV$ c) The electric field is $5.00\times10^{6} V/m$

Work Step by Step

a) The capacitance is given by the formula $C = \frac{ε_{0}A}{d}$. Therefore the capacitance of the capacitor is $$\frac{8.854\times10^{-12}F/m \times 9.82\times10^{-4}m^{2}}{0.00328 m}$$. $$=2.65\times10^{-12}F$$ b) The potential that is stored on a capacitor is given by the equation: $V_{ab} = \frac{Q}{C}$. Therefore the voltage is $\frac{4.35\times10^{-8}C}{2.65\times10^{-12}F}$ $= 16,400 V$ or $16.4 kV$ c) Since the plates are charged there is also a uniform electric field which can be found from this formula: $E=\frac{V}{d}$. Therefore the strength of the electric field is $E = \frac{16400 V}{0.00328 m}$ $=5.00\times10^{6} V/m$
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